\(\begin{vmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta \end{vmatrix}\) = (cos\(\theta\)cos\(\theta\)) - (-sin\(\theta\)sin\(\theta\)) = cos² \(\theta\) + sin²\(\theta\) = 1.

Verify  (B^T)^T = B
Given B = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \) ---(6)
\(\therefore\) B^T = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \)
Also, (B^T)^T = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \) --(7)
From (6) and (7), (B^T)^T = B

Given \(a_{ij}={(i-2j)^2\over 2}with \ m=2,n=3\)
We need to construct a 2 \(\times\) 3 matrix.
\(A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{array}\right]\)
a₁₁ = \(\frac { { (1-2(1)) }^{ 2 } }{ 2 } =\frac { { (-1) }^{ 2 } }{ 2 } =\frac { 1 }{ 2 } \)
a₁₂ = \(\frac { { (1-2(2)) }^{ 2 } }{ 2 } =\frac { { (-3) }^{ 2 } }{ 2 } =\frac { 9 }{ 2 } \)
a₁₃ = \(\frac { { (1-2(3)) }^{ 2 } }{ 2 } =\frac { { (-5) }^{ 2 } }{ 2 } =\frac { 25 }{ 2 } \)
a₂₁ = \(\frac { { (2-2(1)) }^{ 2 } }{ 2 } =\frac { 0 }{ 2 } =0\)
a₂₂ = \(\frac { { (2-2(2)) }^{ 2 } }{ 2 } =\frac { { (-2) }^{ 2 } }{ 2 } =\frac { 4 }{ 2 } \)
a₂₃ = \(\frac { { (2-2(3)) }^{ 2 } }{ 2 } =\frac { { (-4) }^{ 2 } }{ 2 } =\frac { 16 }{ 2 } \)
\(\therefore\) A = \(\left( \begin{matrix} { a }_{ 11 } & a_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \end{matrix} \right) =\left( \begin{matrix} 1/2 & 9/2 & 25/2 \\ 0 & 4/2 & 16/2 \end{matrix} \right) =\frac { 1 }{ 2 } \left( \begin{matrix} 1 & 9 & 25 \\ 0 & 4 & 16 \end{matrix} \right) \)

The number of elements is the product of number of rows and number of columns.
Therefore, we will find all ordered pairs of natural numbers whose product is 12.
Thus, all the possible orders of the matrix are 1 \(\times\) 12, 12 \(\times\) 1, 2 \(\times\)6, 6 \(\times\)2, 3 \(\times\) 4 and 4 \(\times\) 3.
Since 7 is prime, the only possible orders of the matrix are 1 \(\times\) 7 and 7 \(\times\) 1.

The orders of the two matrices are same. Thus by comparing the corresponding elements, we get
\(4 x=x+6\)
\(4 x-x=6\)
\(3 x=6\)
\(x=2\)
\(2 x+y=7\)
Substituting  x = 2
\(2(2)+y=7\)
\(4+y=7\)
\(y=7-4\)
\(y=3\)
\(\therefore x+y=2+3\)
\(x+y=5\)

3A =\(\begin{bmatrix} 12 & 18 & 6 \\ 0 & 3 & 15 \\ 0 & 9 & 6 \end{bmatrix}\)
3A^T =\(\begin{bmatrix} 12 & 0 & 0 \\ 18 & 3 & 9 \\ 6 & 15 & 6 \end{bmatrix}\)= 3\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)= 3(A^T).

A -B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 & 3 \\ -3 & 2 & 1 \\1 & 1 & 1 \end{bmatrix}\)
(A - B)^T =\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\) ..(1)
A^T - B^T =\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\).....(2)
From (1) and (2), (A - B)^T= A^T - B^T.

By the definitions of addition and subtraction of matrices, we have
A+B =\(\begin{bmatrix} 4+\sqrt{3} & 2\sqrt { 5 } & 14.3 \\ 0 & {1\over3} & {3\over4} \end{bmatrix}\) and A - B =\(\begin{bmatrix} 4-\sqrt{3} &0 & -0.3 \\ -2 & -{1\over3} & {1\over4} \end{bmatrix}\)

As the orders of the two matrices are same, they are equal if and only if the corresponding entries are equal. Thus, by comparing the corresponding elements, we get
3x + 4y = 2, x - 2y = 4, a + b = 5, and 2a - b = -5.
Solving these equations, we get x = 2, y = -1, a = 0, and b = 5.

\((i) \text {Let } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]\)
\(A B=\left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} 1+0 & -1+4 \\ 3+0 & -3+8 \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ 3 & 5 \end{array}\right]\)
\(B A=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1-3 & 2-4 \\ 0+6 & 0+8 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 6 & 8 \end{array}\right]\)
\(\therefore A B \neq B A\)
\(\text {(ii)Let } A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \neq 0, B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \neq 0\)
\(A B=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(B A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(\therefore A B=0=B A\)
\(\text { (iii) } A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\)
\(A B=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(B A=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+1 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right] \neq 0\)
\(\therefore A B =0\)
\(B A \neq 0 \)